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匿名 發問於 科學及數學其他 - 科學 · 6 年前

Fluid mechanics

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  • 天同
    Lv 7
    6 年前
    最愛解答

    Consider point A and the nozzle, apply Bernoulli's Equation,

    Pa + (1/2)p(Va)^2 = Pn + (1/2)p(Vn)^2 + pg(1.1)

    where Pa and Pn are the pressures at A and at the nozzle respectively,

    p os the density of water (= 1000 kg/m^3)

    Va and Vn are the speeds of water at A and at the nozzle respectively

    g is the acceleration due to gravity

    Because the water is at free flow at the nozzle, Pn = 0

    thus, (55 x 10^3) + (1000/2)(Va)^2 = (1000/2)(Vn)^2 + 1000 x 9.81 x 1.1

    simplifying, 88.42 + (Va)^2 = (Vn)^2 ----------- (1)

    By equation of continuity,

    (Aa).(Va) = (An).(Vn)

    where Aa and An are the cross-sectional areas of the pipe at point A and at the nozzle respectively.

    hence, [pi(0.2/2)^2].(Va) = [pi.(0.1/2)^2].(Vn)

    0.04(Va) = 0.01(Vn)

    i.e. Va = Vn/4 ----------- (2)

    Substitute Va from (2) into (1),

    88.42 + (Vn/4)^2 = (Vn)^2

    solve for (Vn)^2 gives (Vn)^2 = 94.31 (m/s)^2

    By conservation of mechanical energy,

    gain in potential energy of water at top of water jet = loss of kinetic energy of water at nozzle,

    thus, pgh = (1/2)p(Vn)^2

    h = (Vn)^2/(2g) = 94.31/(2x9.81) m = 4.81 m

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